$\sum_{n \in \mathbb{N}} x_n$ converges if $\sum_{n \in \mathbb{N}} \|x_n\|$ is bounded?

Let $$(x_n)_{n \in \mathbb{N}}$$ be a sequence in a Banach space $$X$$. The series $$\sum_{n \in \mathbb{N}} x_n$$ converges in $$X$$ if $$\sum_{n \in \mathbb{N}} \|x_n\| < \infty$$.

Is this statement true?

Since $$X$$ is a Banach space, if we prove the series $$\sum_{n \in \mathbb{N}} x_n$$ a Cauchy sequence then it converges. That is, for any $$m \ge n > N$$, there is $$\|\sum_{m \in \mathbb{N}} x_m - \sum_{n \in \mathbb{N}} x_n\| = \|\sum_{k = n}^m x_k\| < \epsilon$$ for any $$\epsilon > 0$$. But how can this be derived from the boundedness of $$\sum_{n \in \mathbb{N}} \|x_n\|$$?

• Write the Cauchy condition for the positive series $\sum \|x_n\|$. – Giuseppe Negro Sep 24 '18 at 18:09

Triangle inequality: that follows from the fact that$$\left\lVert\sum_{k=n}^mx_k\right\rVert\leqslant\sum_{k=n}^m\lVert x_k\rVert.$$
• yes but how to show $\sum_{k = n}^m \|x_k\| < \epsilon$ from $\sum_{k = n}^\infty \|x_k\| < \infty$? – Analysis Newbie Sep 24 '18 at 20:10
• The sequence $\left(\sum_{k=1}^n\lVert x_k\rVert\right)_{n\in\mathbb N}$ is increasing and bounded. Therefore, it converges. It follows that it is a Cauchy sequence. – José Carlos Santos Sep 24 '18 at 20:14
In Banach spaces, absolute convergence implies convergence. Since $$s_{n} = \sum_{k=0}^{n}\lVert x_{k}\rVert$$ in monotonic and bounded, it converges. Therefore the given series $$\sum_{n\in\mathbb{N}}x_{n}$$ converges: